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प्रश्न
In the given figure, O is the centre of the circle and B is a point of contact. seg OE ⊥ seg AD, AB = 12, AC = 8, find (1) AD (2) DC (3) DE.
उत्तर
(1) Line AB is the tangent at point B and seg AD is the secant. ...(Given)
∴ By the tangent secant segments theorem,
∴ AC × AD = AB2
∴ 8 × AD = 122
∴ 8 × AD = 144
∴ AD = `144/8`
∴ AD = 18 units
(2) AD = AC + DC ... [A – C – D]
∴ 18 = 8 + DC
∴ DC = 18 – 8
∴ DC = 10 units
(3) seg OE ⊥ seg AD ...[Given]
i.e. seg OE ⊥ seg CD ...[A – C – D]
∴ DE = `1/2 × "DC"` ...(Perpendicular drawn from the centre of the circle to the chord bisects the chord)
∴ DE = `1/2 × 10`
∴ DE = 5 units
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